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先用第一个条件 = gunakan syarat pertama 取整到千要看百位，这个数的百位是 6，一定会进位，千位加 1。结果的千位是 8，所以原来的千位应该是 8 − 1 = 7。可以代进去验证：37 604 → 38 000。 例子：若 5□ 300 取整到千是 52 000，百位 3 不进位，□ 就是 2。 注意：进位的时候，原来的数字比结果小 1。

use the first condition = gunakan syarat pertama Rounding to the nearest thousand depends on the hundreds, which is 6 here, so it always rounds up and the thousands digit goes up by 1. The rounded thousands digit is 8, so the missing digit must be 8 − 1 = 7. Check: 37 604 → 38 000. Example: if 5□ 300 rounds to 52 000, the 3 hundreds do not round up, so □ is 2. Watch out: when it rounds up, the starting digit is 1 less than the rounded one.

用第二个条件检查 = semak dengan syarat kedua 把 7 放回去得到 37 604。取整到万要看千位，千位 7 属于 5 到 9，万位 3 加 1，得到 40 000，和题目一样。两个条件都符合，答案就是 7。 例子：如果检查时结果不同，就要回头换一个数字再试。 注意：答案要让所有条件都成立。

check with the second condition = semak dengan syarat kedua Putting 7 back gives 37 604. Rounding to the nearest ten thousand depends on the thousands digit, 7, which is in 5 to 9, so the ten thousands go from 3 to 4: 40 000, as the question says. Both conditions work, so the answer is 7. Example: if a check gives a different result, go back and try another digit. Watch out: the answer must make every condition true.

推出数字 = cari digit 题目给出两个取整结果，要推出方框里的千位数字。先用“取整到千”的条件：看百位决定千位有没有进位——百位 0 到 4 不进位，千位和结果一样；百位 5 到 9 要进位，原来的千位比结果少 1。推出来以后，再用“取整到万”的条件检查一次。 例子：8□ 150 取整到千是 87 000：百位 1 不进位，□ 就是 7；再检查 87 150 取整到万是 90 000。 注意：推出数字后一定要代回原数，检查所有条件都成立。

find the digit = cari digit Two rounded results are given; find the missing thousands digit. Use the nearest-thousand condition first: the hundreds digit tells you whether the thousands rounded up. With 0 to 4 hundreds the starting thousands digit equals the rounded one; with 5 to 9 it is one less. Then check the nearest-ten-thousand condition. Example: if 8□ 150 rounds to 87 000, the 1 hundred does not round up, so □ is 7; then 87 150 rounds to 90 000. Watch out: put the digit back into the number and check every condition.

连续进位 = kumpul semula berterusan 当一个数后面几位都是 9 时，取整到十、百、千、万，结果可能全都一样。原因是：一进位，9 + 1 = 10，又要再往左进位，一路进上去，后面的数位全变成 0。解这种题时，要把每一个取整条件都代进去检查。 例子：48 996 取整到十 → 个位 6 进位 → 十位 9 + 1 = 10 → 一路进位 → 49 000；取整到百、到千也都是 49 000。 注意：只有连着的 9 才会一路进位；中间有一个不是 9 的数字，进位就会在那里停下来。

regroup all the way = kumpul semula berterusan When the last few digits of a number are all 9, rounding to tens, hundreds, thousands or ten thousands can give the same answer. Rounding up makes 9 + 1 = 10, which carries again to the left, and all the places behind become 0. Check every rounding condition in this kind of puzzle. Example: 48 996 to the nearest ten rounds up, the tens 9 + 1 = 10 keeps carrying, so it becomes 49 000; to the nearest hundred or thousand it is also 49 000. Watch out: the carrying only continues through a row of 9s; a digit that is not 9 stops it.

三个条件同时成立 = tiga syarat serentak 题目说这个数取整到十、到百、到千，三个结果都一样，要找出方框里的数字。做法：把每个可能的数字放进方框，得到一个完整的数；再把这个数分别取整到十、到百、到千，算三次；三个结果都等于题目给的数，这个数字才对。 例子：检查 25 996：取整到十、到百、到千都得 26 000，三个结果一样；换成 25 796，取整到十只得 25 800，结果就不一样了。 注意：三个条件都要真的算一次；只检查其中一个条件，常常会以为好几个数字都可以。

three conditions at once = tiga syarat serentak The number rounds to the same value to the nearest ten, hundred and thousand; find the missing digit. Put each possible digit into the box, round the whole number three times, and keep the digit only if all three results match the given number. Example: check 25 996: to the nearest ten, hundred and thousand it gives 26 000 each time; but 25 796 to the nearest ten gives only 25 800. Watch out: really work out all three roundings; checking only one condition makes several digits look possible.